L'interaction du mix

1.                               # Modèle ou l`elasticité prix est fonction de l`effort publicitaire

2.                               x1<-seq(0,5,0.5)

3.                               x2<-rep(11,11)

4.                               f1 <- 0.2+2.3*x1^1.5/(3.5^1.5+x1^1.5) # pub 5.2

5.                               f2 <- 5*x2^(-0.25*(x1-4)) # prix 5.3

6.                               y<-100*f1*f2 # 5.1

7.                               profit<-(x2-3)*y-x1

8.                               df<-data.frame(Pub=x1, Ventes=y, Profit=profit)

9.                               f2 <- 5*x2^(-0.25*(1.5-4)) # elasticité du prix fixe pour une pub de 1.5

10.                             y<-100*f1*f2 # 5.1

11.                             profit<-(x2-3)*y-x1

12.                             df$Profit2=profit

13.                             df

14.                             matplot(x1, df[,3:4], pch = 1:2, type = "o", col = 1:2,xlab="Valeurs de x", ylab="Ventes et/ou Profits")

15.                             legend(min(x1), max(df[,3:4]),names(df)[3:4], lwd=3, col=1:2, pch=1:2)

1.                               # Modèle de l`usure de la publicité

2.                               a=10

3.                               b=10

4.                               c=-1

5.                               t<-1:10

6.                               bt<-b*exp(-0.07*(t-1)) # effet d`usure sur le coef b de la pub

7.                               xopt<-(1-0.3*b)/(0.6*c) # pub optimum ingorant l`usure

8.                               x<-rep(xopt,10)

9.                               y<-a+bt*x+c*x^2

10.                             profit<-0.3*y-x

11.                             df<-data.frame(Temps=t, Profit1=profit)

12.                             xopt_t<-(1-0.3*bt)/(0.6*c) # pub optimum utlisant l`usure

13.                             x<-xopt_t

14.                             profit<-0.3*y-x

15.                             df$Profit2=profit

16.                             df

17.                             matplot(t, df[,2:3], pch = 1:2, type = "o", col = 1:2,xlab="Valeurs de x", ylab="Ventes et/ou Profits")

18.                             legend(mean(t), max(df[,2:3]),names(df)[2:3], lwd=3, col=1:2, pch=1:2)

1.                               # Variation du budget optimum quand l interaction des variables du mix et positive, negative ou zero

2.                               a=50

3.                               b=200

4.                               c=2

5.                               d=2

6.                               x1<-seq(6,7.9,0.1)

7.                               x2<-rep(7,20)

8.                               f1<-a+(b-a)*x1^c/(d^c+x1^c)

9.                               f2<-a+(b-a)*x2^c/(d^c+x2^c)

10.                             y<-f1+f2 # sans interaction

11.                             profit<-0.3*y-x1-x2

12.                             df<-data.frame(Effort=x1, Profit.InteractZero=profit)

13.                             y<-f1+f2+0.001*f1*f2 # interaction positive

14.                             profit<-0.3*y-x1-x2

15.                             df$Profit.InteractPos<-profit

16.                             y<-f1+f2-0.001*f1*f2 # interaction negative

17.                             profit<-0.3*y-x1-x2

18.                             df$Profit.InteractNeg<-profit

19.                             df

20.                             matplot(x1, df[,2:4], pch = 1:2, type = "o", col = 1:3,xlab="Valeurs de x", ylab="Ventes et/ou Profits")

21.                             legend(0.75*max(x1), max(df[,2:4]),names(df)[2:4], lwd=3, col=1:3, pch=1:3)